Question

A mixture of \(1.00 \mathrm{mol} \mathrm{NaHCO}_{3}(\mathrm{s})\) and \(1.00 \mathrm{mol}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is introduced into a \(2.50 \mathrm{L}\) flask in which the partial pressure of \(\mathrm{CO}_{2}\) is 2.10 atm and that of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is \(715 \mathrm{mmHg} .\) When equilibrium is established at \(100^{\circ} \mathrm{C},\) will the partial pressures of \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) be greater or less than their initial partial pressures? Explain. $$\begin{array}{r} 2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}}=0.23 \mathrm{at} 100^{\circ} \mathrm{C} \end{array}$$

Short answer

The partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}O\) will be less than their initial values because in order to reach equilibrium, the reaction shifts to the left, consuming \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}O\).

Step-by-step solution

Understand the Equation

The chemical equation is: \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). At \(100^{\circ} \mathrm{C}\), the equilibrium constant \(K_{\mathrm{p}}\) is 0.23.

Convert Pressure Units

The initial pressure of \(\mathrm{CO}_{2}\) is given as 2.10 atm, while that of \(\mathrm{H}_{2}O\) is given as 715 mmHg. Convert \(\mathrm{H}_{2}O\) pressure to atm by dividing by 760 (since 760 mmHg is equivalent to 1 atm), getting approximately 0.941 atm.

Calculate Reaction Quotient Q

The reaction quotient \(Q_{\mathrm{p}}\) is calculated using the initial conditions before the system reaches equilibrium. Here, \(Q_{\mathrm{p}} = [\mathrm{CO}_{2}][\mathrm{H}_{2}O]\) (since the stoichiometric coefficients in the balanced equation are all 1 and the reactants are in solid state, so they aren't included in the expression). So, \(Q_{\mathrm{p}} = (2.1)(0.941) = 1.98\).

Compare Q with Kp

Now, compare \(Q_{\mathrm{p}}\) (1.98) with \(K_{\mathrm{p}}\) (0.23). Since \(Q_{\mathrm{p}} > K_{\mathrm{p}}\), the system will shift to the left (towards the reactants) to reach equilibrium. This means that the number of moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}O\) will decrease, thus lowering their partial pressures.

Key concepts

Equilibrium Constant (Kp)

Understanding the equilibrium constant, denoted as \( K_p \), is crucial for students delving into the behavior of gases in chemical reactions that have reached a state of balance. The equilibrium constant for a reaction involving gases is expressed in terms of partial pressures and is specific to a given temperature, indicating the ratio of the products to reactants at equilibrium.

When the equilibrium constant is known, like the \( K_p = 0.23 \) in our exercise, it tells us how far a reaction will proceed before reaching equilibrium. A smaller \( K_p \) value implies fewer products are present at equilibrium relative to the reactants. In contrast, a larger \( K_p \) suggests a reaction that favors the formation of products.

When solving problems related to \( K_p \), remember that it only includes gases and not solids or liquids. Hence, for the chemical reaction in the exercise, the solid reactants do not appear in the expression for \( K_p \).

Reaction Quotient (Qp)

The reaction quotient, \( Q_p \), serves as a 'snapshot' of a reaction's status before equilibrium is reached. It is calculated in the same way as the equilibrium constant, \( K_p \), using the partial pressures of the gases involved.

In our exercise, the calculation of the reaction quotient at the start shows a \( Q_p = 1.98 \), comparing the partial pressures of \( CO_2 \) and \( H_2O \) before the system reaches equilibrium. This number doesn't just exist in isolation; its value is significant only when compared to the equilibrium constant, as it indicates in which direction the reaction must shift to achieve equilibrium. Remember, the concentrations (or pressures) used to calculate \( Q_p \) are the initial ones, not those at equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a guiding concept in understanding how a chemical system at equilibrium reacts to changes in concentration, temperature, or pressure. According to this principle, if an external stress is applied to a system at equilibrium, the system adjusts to minimize that stress and restore a state of balance.

In the exercise, because \( Q_p > K_p \), Le Chatelier's principle tells us that the reaction will shift to the left to reduce the concentrations of the products (in this case, gases) and increase the concentration of reactants. Essentially, it means that an excess of products causes the reaction to favor the formation of reactants to re-establish equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It's one part of the total pressure exerted by the mixture, which is equal to the sum of the partial pressures of each individual gas (Dalton's Law).

In our exercise, we initially have the partial pressures of \( CO_2 \) and \( H_2O \), which are crucial to calculating the reaction quotient. To correctly analyze the system, we convert the pressure of \( H_2O \) to the same units as \( CO_2 \) (atmospheres) to maintain consistency in our calculations. This uniformity is important in stoichiometry and equilibrium calculations involving gases.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants used and products formed by a chemical reaction, based on the balanced chemical equation. It's a fundamental concept in chemistry that allows us to predict the amounts of substances consumed and produced.

In the context of our exercise, stoichiometry helps us understand the mole-to-mole relationships outlined in the balanced equation. Notably, these stoichiometric coefficients become crucial when dealing with equilibrium and reaction quotient expressions as they dictate the ratio of reactants to products in these calculations. However, remember that for the purposes of equilibrium expressions, only the gaseous substances are accounted for.